Answer:
The mixture will be entirely solid.
Explanation:
Mass of solid gold: ms=2.0kg Solid Initial tremperature: Tos= 727°C (Absorbs heat)
Mass of liquid gold: ml=1.5kg Liquid Initial tremperature: Tol= 1063°C (Release heat)
Two bodies reach thermal equilibrium when the heat transmitted is equal to the heat absorbed and both reach the same temperature (equilibrium temperature):
Q= Heat
CeAu= Specific heat of gold
Q absorbed = Q released
Te= equilibrium temperature
ms*CeAu*(Te-Tos)= ml*CeAu*(Tol-Te)
You can simplify the specific heat of gold, and we have:
ms*(Te-Tos)= ml*(Tol-Te)
Applying distributive property:
ms*Te - ms*Tos = ml*Tol - ml*Te
We place the terms "Te" on the right side of equality and on the left side of equality the known data:
ms*Te + ml*Te = ml*Tol + ms*Tos
"Te" as a common factor:
Te*(ms +ml) = ml*Tol + ms*Tos
Te = (ml*Tol + ms*Tos)/ (ms +ml)
Te= (1.5kg*1063°C + 2.kg *727°C)/(2.0kg+1.5kg)
Te= 3,048.5kg*°C/3.5kg
Te= 871°C
melting point of gold = 1,064 °C
At 871°C gold is solid
